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Blog - Exponential Zero (changes)

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This page is a blog article in progress, written by David Tanzer. To see discussions of this page as it was being written, go to the Azimuth Forum.

Here is a mathematical riddle. Consider the function below, which is undefined for negative values, sends zero to one, and sends positive values to zero. Can you come up with a nice compact formula for this function, which uses only the basic arithmetic operations, such as addition, division and exponentiation? You can’t use any special functions, including things like sign and step functions, which are by definition discontinuous.

In college, I ran around showing people the graph, asking them to guess the formula. I even tried it out on some professors there, U. Penn. My algebra prof, who was kind of intimidating, looked at it, got puzzled, and then got irritated. When I showed him the answer, he barked out: Is this exam over??! Then I tried it out during office hours on E. Calabi, who was teaching undergraduate differential geometry. With a twinkle in his eye, he said, why that’s zero to the x!

The graph of 0 x0^x is not without controversy. It is reasonable that for positive xx, we have that 0 x0^x is zero. Then 0 x=1/0 x=1/00^{-x} = 1/0^x = 1/0 is undefined – so the the function is undefined for negative values. But what about 0 00^0? This question is bound to come up in the course of one’s general mathematical education, and has been the source of long, ruminative arguments.

There are three contenders for 0 00^0: undefined, 0, and 1. Let’s try to define it in a way that is most consistent with the general laws of exponents. In particular: for all a,x,ya,x,y, a x+y=a xa ya^{x+y} = a^x a^y, and a x=1/a xa^{-x} = 1/a^x. Hence:

  • 0 0=0 0+0=0 00 00^0 = 0^{0 + 0} = 0^0 0^0
  • 0 0=0 0=1/0 00^0 = 0^{-0} = 1/0^0

So 0 00^0 equals its square, and equals its reciprocal. By these criteria, it follows that 0 00^0 is 1.

That is the justification for the above graph – and for the striking discontinuity that it contains.

Here is an intuitive way to understand the discontinuity. Consider the family of exponential curves b xb^x, with bb as the parameter. When b=1b = 1, you get the constant function 1. When bb is more than 1, you get an increasing exponential, and when it is between 0 and 1, you get a decreasing exponential. The intersection of all of these graphs is the “pivot” point x=0x = 0, y=1y = 1. That is the “dot” of discontinuity.

Now what happens to b x\b^x, in the limit, as bb decreases to zero? To the right of the origin, the curve progressively flattens down to zero. To the left it rises up towards infinity faster and faster. But it always “crosses” through the point x=0x = 0, y=1y = 1, so this point remains in the limiting curve. In heuristic terms, the value y=1y = 1 is the discontinuous transition from infinitesimal values to infinite values.

There are reasons, however, why the value 0 00^0 can be considered as indeterminate, and left undefined. These were indicated by the good professor.

Dr. Calabi had a truly inspiring teaching style, back in the day. He spoke of Italian paintings, and showed a kind of geometric laser vision.  In the classroom, he showed us the idea of torsion using his arms to fly around the room like an airplane.  There’s even a manifold named after him, the Calabi-Yau manifold.

He went on to talk about the underpinnings of this quirky function. He called attention to the function f(x,y)=x yf(x,y) = x^y, over the complex domain, and attempted to sketch its level sets. He focused on the behavior of the function when xx and yy are close to zero. Then he stated that every one of the level sets L(z)={(x,y)|x y=z}L(z) = \{(x,y)|x^y = z\} comes arbitrarily close to (0,0)(0,0).

This means that x yx^y has a wild singularity at the origin: every complex number zz is the limit of x yx^y along some path to zero. Indeed, to reach zz, just take a path in L(z)L(z) that approaches (0,0)(0,0).

To see why the level sets all approach the origin, take logs, to get ln(x y)=xln(y)=ln(z)ln(x^y) = x ln(y) = ln(z). That gives y=z/ln(x)y = z / ln(x), which is a parametric formula for L(z)L(z). As xx goes to zero, ln(x)ln(x) goes to -\infinity, and so yy goes to zero. These are paths (x,z/ln(x))(x, z/ln(x)), completely within L(z)L(z), which approach the origin.

In making these statements, we need to bear in mind that x yx^y is multi-valued. That’s because x ye ln(x y)=e yln(x)x^y \equiv e^{ln(x^y)} = e ^ {y ln(x)}, and the natural logarithm ln(x)ln(x) is multi-valued. And that is because ln(x)ln(x) is the inverse of the complex exponential, which is many-to-one: adding any integer multiple of 2πi2 \pi i to zz leaves e ze^z unchanged. And that follows from the definition of the exponential, which sends a+bia + bi to the complex number with magnitude aa and phase bb.

Footnote: to visualize these operations, represent the complex numbers \mathbb{C} by the real plane 2\mathbb{R}^2. Addition in \mathbb{C} is given by vector addition in 2\mathbb{R}^2. Multiplication gives the vector with magnitude equal to the product of the magnitudes, and phase equal to the sum of the phases. The positive real numbers have phase zero, and the positive imaginary numbers are at 90 degrees vertical, with phase π/2\pi / 2.

How many values does x yx^y have? Well, ln(x)ln(x) has a countable number of values, all differing by integer multiples of 2πi2 \pi i. This generally induces a countable number of values for x yx ^y. But if yy is rational, they collapse down to a finite set. When y=1/ny = 1/n, for example, the values of yln(x)y ln(x) are spaced apart by (2πi)/n(2 \pi i) / n, and when these get pumped back through the exponential function, we find only nn distinct values for x 1/nx ^ {1/n}. These are the nth roots of xx.

So, to speak of the limit of x yx^y along a path, and of the partition of 2\mathbb{C}^2 into level sets, we need to work within a branch of x yx^y. Each branch induces a different partition of 2\mathbb{C}^2. But for all of these partitions, it holds true that all of the level sets approach the origin. That follows from the formula for the level set L(z)L(z), which is y=z/ln(x)y = z / ln(x). As xx goes to zero, every branch of ln(x)ln(x) goes to -\infinity. (Exercise: why?) Hence yy also goes to zero. The branch affects the shape of the paths to the origin, but not their existence.

Here is a qualitative description of how the level sets fit together: they are like spokes around the origin, where each spoke is a curve in one complex dimension. These curves are 1-D complex manifolds, which are equivalent to two-dimensional surfaces in 4\mathbb{R}^4. The partition comprises a two-parameter family of these surfaces, indexed by the complex value of x yx^y.

What can be said about the geometry and topology of this “wheel of manifolds”? We know they don’t intersect. But are they “nicely” layered, or twisted and entangled? As we zoom in on the origin, does the picture look smooth, or does it have a chaotic appearance, with infinite fine detail? Suggestive of chaos is the fact that the gradient x y=(yx y1,ln(y)x y)=(y/x,ln(y))x y\nabla x^y = (y x^{y-1}, ln(y) x^y) = (y/x, ln(y)) x^y is also wildly singular at the origin.

These questions can be explored with plotting software. Here, the artist would have the challenge of having only two dimensions to work with, when the “wheel” is really a structure in four-dimensional space. So some interesting cross-sections would have to be chosen.

Exercises:

  • Speak about the function b xb^x, where bb is negative, and xx is real.

  • What is 0 π0^\pi, and why?

  • What is 0 i0^i?

Moral: something that seems odd, or like a joke that might annoy your algebra prof, could be more significant than you think. So tell these riddles to your professors, while they are still around.